A particle of mass 1 kg is free to move alon | vedclass

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(B) The potential energy is given by $U(x) = \frac{x^2}{2} - x$. To find the minimum potential energy,we set the derivative to zero: $\frac{dU}{dx} =

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$\sqrt{5} \ ms^{-1}$

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(B) The potential energy is given by $U(x) = \frac{x^2}{2} - x$. To find the minimum potential energy,we set the derivative to zero: $\frac{dU}{dx} = x - 1 = 0$,which gives $x = 1 \ m$. The second derivative $\frac{d^2U}{dx^2} = 1 > 0$,confirming a minimum at $x = 1 \ m$. The minimum potential energy is $U_{min} = \frac{(1)^2}{2} - 1 = -0.5 \ J$. The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U$. Since $E = 2 \ J$,the maximum kinetic energy $K_{max}$ occurs when $U$ is at its minimum: $K_{max} = E - U_{min} = 2 - (-0.5) = 2.5 \ J$. Using $K_{max} = \frac{1}{2}mv_{max}^2$,we have $2.5 = \frac{1}{2}(1)v_{max}^2$. $v_{max}^2 = 5$,so $v_{max} = \sqrt{5} \ ms^{-1}$.

$A$ particle of mass $1 \ kg$ is free to move along the $x$-axis. Its potential energy is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the particle is $2 \ J$,find the maximum speed of the particle.

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